# 477. Total Hamming Distance (Medium)

https://leetcode.com/problems/total-hamming-distance/

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Now your job is to find the total Hamming distance between all pairs of the given numbers.

Example:

```Input: 4, 14, 2

Output: 6

Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
showing the four bits relevant in this case). So the answer will be:
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
```

Note:

1. Elements of the given array are in the range of `0 ` to `10^9`
2. Length of the array will not exceed `10^4`.

## Solutions

``````class Solution {
public int totalHammingDistance(int[] nums) {
int sum = 0;
for (int i = 0; i < 32; i++) {
int numOfOnes = 0;
int p = (1 << i);

for (int num : nums) {
if ((num & p) > 0) {
numOfOnes++;
}
}

// The total size of combination pairs will be (nums.length - 1) * nums.length / 2.
// You don't have to go through all the combinations to gain the result.

// The trick is to figure out how many 0s and 1s on a specific bit since only
// either pair (0,1) or (1,0) will count for the distance calculation.
// Then you calculate the total combinations of 0s and 1s and derive the the total distance.

// The total size of combination pairs of 0 and 1 will be ((nums.length - numOfOnes) * numOfOnes)
sum += ((nums.length - numOfOnes) * numOfOnes);
}

return sum;
}
}
``````